Question

A capillary tube of 0.4 mm diameter is placed vertically inside (i) water of surface tension 6.5 x 10-2 Nm-1 and zero angle of contact, (ii) a liquid of density 800 kgm-3, surface tension 5.0 x 10-2 Nm-1 and an angle of contact 30 degrees. Calculate the height to which the liquid rises in the capillary in each case.

Answer 1

Let $r$ be the radius of the capillary tube, $h$ the rise of the liquid in the capillary, $\theta$ the angle of contact.

The liquid makes contact with the tube along a line of length $2\pi r$ . As the surface tension $S$ acts tangentially to the liquid surface, its vertical component is $S \cos\theta$ so that the total upward force due to it is $2\pi r S\cos\theta$ . This balances the weight of a liquid cylinder of height $h$ and weight $\pi r^2h\rho g$ , where $\rho$ is the density of the liquid.

$\therefore 2\pi rS\cos\theta=\pi r^2h\rho g\\ \Rightarrow h=\frac{2S\cos\theta}{r\rho g}\qquad ...........(1)$

Diameter of the capillary tube = 0.4 mm

$\therefore$ Radius of the capillary tube, $r=0.2\ mm = 0.2\times 10^{-3}\ m$

(i) For water:

Surface tension of water, $S=6.5\times 10^{-2}\ N/m$

Contact angle, $\theta=0\degree$

Density of water, $\rho = 1000\ kg/m^3$

Substituting the values in Eq.(1),

$h=\frac{2\times 6.5\times 10^{-2}\times \cos0\degree}{0.2\times 10^{-3}\times 1000\times 9.8}\ m\\ \Rightarrow h=6.6\times 10^{-2}\ m\\ \Rightarrow h=6.6\ cm$

(ii) For the liquid:

Surface tension, $S=5.0\times 10^{-2}N/m$

Contact angle, $\theta=30\degree$

Density of the liquid, $\rho=800\ kg/m^3$

$\therefore h=\frac{2\times 5.0\times 10^{-2}\times \cos30\degree}{0.2\times 10^{-3}\times 800\times 9.8}\ m\\ \Rightarrow h=5.5\times 10^{-2}\ m\\ \Rightarrow h=5.5\ cm$

Answer: (i) Water rises to 6.6 cm in the capillary and (ii) the liquid rises to 5.5 cm